15_R_3 - y = 1), then ZZ x dS = √ 2 Z 1-1 dy Z 1-y 2 x dx...

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3. The cone z = p x 2 + y 2 has area element dS = s 1 + x 2 + y 2 z 2 dx dy = 2 dx dy . If is the part of the cone in the region 0 x 1 - y 2 (which itself lies between y = - 1 and
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Unformatted text preview: y = 1), then ZZ x dS = √ 2 Z 1-1 dy Z 1-y 2 x dx = 2 √ 2 Z 1 1-2 y 2 + y 4 2 dy = 8 √ 2 15 ....
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This note was uploaded on 03/08/2012 for the course MATH 120 taught by Professor Onurfen during the Spring '12 term at Middle East Technical University.

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